Electrical Machine - IES Academy
Electrical Machine - IES Academy Electrical Machine - IES Academy
India’s No. 1 IES Academy Transformer Chapter-1 (a) Voltage-drop in the primary impedance due to the secondary current (b) Voltage-drop in the primary impedance due to the exciting current (c) Voltage-drop in the secondary impedance due to the exciting current (d) Reduction in values of R C and XQof the exciting circuit Other Questions IES-56. Consider the following statements regarding transformers: [IES-2007] 1. The function of the magnetizing component of no load current is to sustain the alternating flux in the core. 2. Short circuit test is performed to find core losses only. 3. The function of the breather in transformer is to arrest flow of moisture when outside air enters the transformer. Which of the statements given above are correct′ (a) 1 and 2 (b) 1 and 3 (c) 2 and 3 (d) 1, 2 and 3 IES-57. IES-58. Sludge formation in transformer oil is due to which one of the following (a) Ingress of duct particles and moisture in the oil [IES-2008] (b) Appearance of small fragments of paper, varnish, cotton and other organic materials in the oil (c) Chemical reaction of transformer oil with the insulating materials (d) Oxidation of transformer oil Match List I (Transformer) with List II (Voltage ratio) and select the correct answer: [IES-2002] List-I List–II A. Power transformer 1. 230V/230 V B. Auto transformer 2. 220V/240 V C. Welding transformer 3. 400V/100V D. Isolation transformer 4. 132kV/11kV Codes: A B C D A B C D (a) 4 2 3 1 (b) 4 2 1 3 (c) 2 4 1 3 (d) 2 4 3 1 www.iesacademy.com E-mail: iesacademy@yahoo.com Page-46 25, 1 st Floor, Jia Sarai, Near IIT. New Delhi-16 Ph: 011-26537570, 9810958290
India’s No. 1 IES Academy Transformer Chapter-1 Answers with Explanation (Objective) Previous Year GATE Answer x× MVA× p.. f GATE-1. Ans. (d) Efficiency = η = 2 x × MVA × p f + W × x + W Where x = % of loading of the transformer WCu = Cu losses at fL condition Wi = iron loss of transformer WCu where ( ηmax ) → then x = = 1 Wi ∴ W Cu = Wi MVA×1×1 ∴ 90% = MVA+2W i ∴ W i = 0.0555 MVA At half load .. Cu i MVA×0.5 0.5 η = = = 87.8% MVA ×0.5 +0.01389 MVA +0.055 MVA 0.5 +0.0694 GATE-2. Ans. (b) x = Wi WCu Wi = Iron loss of transformer WCu = Cu loss of transformer (KVA)0.8 GATE-3. Ans. (c) Efficiency, η = 95% = (KVA)0.8+w +w ........(i) 0.5(KVA) 96% = ........(ii) 0.5(KVA)+0.25w +w cu i cu i GATE-4. Ans. (b) From,equation (i) wcu + wi = 12.63 From equation (ii) 0.25 wcu + 0.96wi = 6.25 From above two equations w = 8.51,w = 4.12 cu 4.12 x = 0.696 8.51 = i www.iesacademy.com E-mail: iesacademy@yahoo.com Page-47 25, 1 st Floor, Jia Sarai, Near IIT. New Delhi-16 Ph: 011-26537570, 9810958290
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India’s No. 1<br />
<strong>IES</strong> <strong>Academy</strong><br />
Transformer<br />
Chapter-1<br />
Answers with Explanation (Objective)<br />
Previous Year GATE Answer<br />
x× MVA×<br />
p..<br />
f<br />
GATE-1. Ans. (d) Efficiency = η =<br />
2<br />
x × MVA × p f + W × x + W<br />
Where x = % of loading of the transformer<br />
WCu = Cu losses at fL condition<br />
Wi = iron loss of transformer<br />
WCu<br />
where ( ηmax<br />
) → then x = = 1<br />
Wi<br />
∴<br />
W<br />
Cu<br />
= Wi<br />
MVA×1×1<br />
∴<br />
90% = MVA+2W<br />
i<br />
∴<br />
W<br />
i<br />
= 0.0555 MVA<br />
At half load<br />
..<br />
Cu i<br />
MVA×0.5 0.5<br />
η = = = 87.8%<br />
MVA ×0.5 +0.01389 MVA +0.055 MVA 0.5 +0.0694<br />
GATE-2. Ans. (b) x =<br />
Wi<br />
WCu<br />
Wi<br />
= Iron loss of transformer<br />
WCu<br />
= Cu loss of transformer<br />
(KVA)0.8<br />
GATE-3. Ans. (c) Efficiency, η = 95% =<br />
(KVA)0.8+w +w<br />
........(i)<br />
0.5(KVA)<br />
96% = ........(ii)<br />
0.5(KVA)+0.25w +w<br />
cu<br />
i<br />
cu<br />
i<br />
GATE-4. Ans. (b)<br />
From,equation (i)<br />
wcu<br />
+ wi<br />
= 12.63<br />
From equation (ii)<br />
0.25 wcu<br />
+ 0.96wi<br />
= 6.25<br />
From above two equations<br />
w = 8.51,w = 4.12<br />
cu<br />
4.12<br />
x = 0.696<br />
8.51 =<br />
i<br />
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